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Abbreviated Solutions to Problems Abbreviated Solutions to Problems Fuller solutions to all chapter problems are published in the Absolute molecules also differ in size and shape—either of these property Ultimate Study Guide to Accompany Principles of Biochemistry. differences could be the basis for separation. (b) Glucose is a For all numerical problems, answers are expressed with the correct smaller molecule than a nucleotide; separation could be based on number of signifi cant fi gures. size. The nitrogenous base and/or the phosphate group also endows nucleotides with characteristics (solubility, charge) that Chapter 1 could be used for separation from glucose. 1. (a) Diameter of magnified cell 5 500 mm (b) 2.7 3 1012 actin 10. It is improbable that silicon could serve as the central organizing 10 molecules (c) 36,000 mitochondria (d) 3.9 3 10 glucose element for life, especially in an O2-containing atmosphere such molecules (e) 50 glucose molecules per hexokinase molecule as that of Earth. Long chains of silicon atoms are not readily 2. (a) 1 3 10212 g 5 1 pg (b) 10% (c) 5% synthesized; the polymeric macromolecules necessary for more 3. (a) 1.6 mm; 800 times longer than the cell; DNA must be tightly complex functions would not readily form. Oxygen disrupts coiled. (b) 4,000 proteins bonds between silicon atoms, and silicon–oxygen bonds are extremely stable and difficult to break, preventing the breaking 4. (a) Metabolic rate is limited by diffusion, which is limited by and making of bonds that is essential to life processes. surface area. (b) 12 ␮mϪ1 for the bacterium; 0.04 ␮mϪ1 for the amoeba; surface-to-volume ratio 300 times higher in the 11. Only one enantiomer of the drug was physiologically active. bacterium. Dexedrine consisted of the single enantiomer; Benzedrine consisted of a racemic mixture. 5. 2 3 106 s (about 23 days) 12. (a) 3 Phosphoric acid groups; ␣-D-ribose; guanine (b) Choline; 6. The vitamin molecules from the two sources are identical; the phosphoric acid; glycerol; oleic acid; palmitic acid (c) Tyrosine; body cannot distinguish the source; only associated impurities 2 glycines; phenylalanine; methionine might vary with the source. 13. (a) CH O; C H O 7. 2 3 6 3 (b) (a) H HH(b) OH H OH ϩ H OH OH H N C C OH HCOH 3 H C C H C C H C C H H HCOHHydroxyls H C OH H C OH H C OH Amino Hydroxyl H C OH HO HO H H 123 OH OH H OH (c) O (d) COOϪ Carboxyl OH H O ϩ HO C C HO C C HCCC HO P OϪ Phosphoryl Amino HC H3N H C H H C OH H H OH O HCOH Hydroxyl H H H C C COOϪ CH 3 456 H Carboxyl Methyl HO H O HO OH O H OH O (e) ϪO O (f) H O HCCC HCCC HCCC Carboxyl Aldehyde C C H H OH H H H H OH H ϩ CH CH NH Amino 2 3 789 CH2 Hydroxyl HO C H NH H C OH HO H HO O OH H O OH Amido O C O H C OH Hydroxyls HCC C H C C HC H C C C H HO H H H H H OH H C OH Hydroxyl CH2OH Methyl H3C C CH3 Methyl 10 11 12 CH2OH Hydroxyl (c) X contains a chiral center; eliminates all but 6 and 8. (d) X contains an acidic functional group; eliminates 8; structure 6 is 8. OH H H consistent with all data. (e) Structure 6; we cannot distinguish between the two possible enantiomers. HO C* CH N C CH 2 3 14. The compound shown is (R)-propranolol; the carbon bearing the H CH HO 3 hydroxyl group is the chiral carbon. (S)-Propranolol has the structure: The two enantiomers have different interactions with a chiral biological “receptor” (a protein). 9. (a) Only the amino acids have amino groups; separation could be ON* based on the charge or binding affinity of these groups. Fatty acids H OH are less soluble in water than amino acids, and the two types of AS-1 AS-2 Abbreviated Solutions to Problems 15. The compound shown is (S,S)-methylphenidate. (R,R)- H Br OH N C H methylphenidate has the structure: H + 9 17 H N O Br H HO O Br HN H O HO H H O N O OH O Br 2 * NH H OH HO O * H H Tetrabromosucrose Sucronic acid (d) First, each molecule has multiple AH-B groups, so it is difficult to know which is the important one. Second, because The chiral carbons are indicated with asterisks. the AH-B motif is very simple, many nonsweet molecules will have this group. (e) Sucrose and deoxysucrose. Deoxysucrose 16. (a) A more negative DG8 corresponds to a larger K for the eq lacks one of the AH-B groups present in sucrose and has a binding reaction, so the equilibrium is shifted more toward slightly lower MRS than sucrose—as is expected if the AH-B products and tighter binding—and thus greater sweetness and groups are important for sweetness. (f) There are many such higher MRS. (b) Animal-based sweetness assays are time- examples; here are a few: (1) D-Tryptophan and 6-chloro-D- consuming. A computer program to predict sweetness, even if tryptophan have the same AH-B group but very different MRS not always completely accurate, would allow chemists to design values. (2) Aspartame and neotame have the same AH-B groups effective sweeteners much faster. Candidate molecules could but very different MRS values. (3) Neotame has two AH-B then be tested in the conventional assay. (c) The range 0.25 to groups and alitame has three, yet neotame is more than five 0.4 nm corresponds to about 1.5 to 2.5 single-bond lengths. The times sweeter than alitame. (4) Bromine is less electronegative figure below can be used to construct an approximate ruler; any than oxygen and thus is expected to weaken an AH-B group, yet atoms in the light red rectangle are between 0.25 and 0.4 nm tetrabromosucrose is much sweeter than sucrose. (g) Given from the origin of the ruler. enough “tweaking” of parameters, any model can be made to fit a defined dataset. Because the objective was to create a model to predict DG8 for molecules not tested in vivo, the researchers There are many possible AH-B groups in the molecules; a few needed to show that the model worked well for molecules it had are shown here. not been trained on. The degree of inaccuracy with the test set could give researchers an idea of how the model would behave H OH for novel molecules. (h) MRS is related to Keq, which is related H exponentially to DG8, so adding a constant amount to DG8 O HO H H O OH multiplies the MRS by a constant amount. Based on the values H H given with the structures, a change in DG8 of 1.3 kcal/mol HO H OH OH O corresponds to a 10-fold change in MRS. H OH H Deoxysucrose Chapter 2 1. Ethanol is polar; ethane is not. The ethanol OOH group can H OH hydrogen-bond with water. H O HO H HO O OH 2. (a) 4.76 (b) 9.19 (c) 4.0 (d) 4.82 H Ϫ4 Ϫ7 Ϫ12 HO H H 3. (a) 1.51 3 10 M (b) 3.02 3 10 M (c) 7.76 3 10 M OH O OH H OH 4. 1.1 H 5. (a) HCl Δ H1 1 Cl2 (b) 3.3 (c) NaOH Δ Na1 1 OH2 Sucrose (d) 9.8 NH2 O 6. 1.1 O H O N 7. 1.7 3 10Ϫ9 mol of acetylcholine O OH OH S 8. 0.1 M HCl NH O O O NH2 9. (a) greater (b) higher (c) lower N 10. 3.3 mL H O CH3 11. (a) RCOOϪ (b) RNH (c) H POϪ (d) HCOϪ D-Tryptophan Saccharin Aspartame 2 2 4 3 12. (a) 5.06 (b) 4.28 (c) 5.46 (d) 4.76 (e) 3.76 O 13. (a) 0.1 M HCl (b) 0.1 M NaOH (c) 0.1 M NaOH NH2 O Ϫ OH H H 14. (d) Bicarbonate, a weak base, titrates OOH to OO , making N N S OH the compound more polar and more water-soluble. NH2 15. Stomach; the neutral form of aspirin present at the lower pH is Cl N O O H less polar and passes through the membrane more easily. 6-Chloro-D-tryptophan Alitame 16. 9 17. 7.4 18. (a) pH 8.6 to 10.6 (b) 4/5 (c) 10 mL (d) pH 5 pKa 2 2 NH O H 19. 8.9 N OH 20. 2.4 O 21. 6.9 O O 22. 1.4 CH3 23. NaH2PO4 ? H2O, 5.8 g/L; Na2HPO4, 8.2 g/L Ϫ Neotame 24. [A ]/[HA] 5 0.10 AbbreviatedSolutionsToProblems.indd Page AS-3 05/10/12 10:59 AM user-F408 /Users/user-F408/Desktop Abbreviated Solutions to Problems AS-3 25. Mix 150 mL of 0.10 M sodium acetate and 850 mL of 0.10 M 4. (a)–(c) acetic acid. COOH 26. Acetic acid; its pK is closest to the desired pH. HN ϩ pK ϭ 1.82 a CH CH NH 1 27. (a) 4.6 (b) 0.1 pH unit (c) 4 pH units ϩ 2 3 N Hϩ 28.
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